Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $x = \dfrac{4y - 24}{-y^2 - y + 42} \times \dfrac{2y^2 + 14y + 12}{y + 1} $
Answer: First factor out any common factors. $x = \dfrac{4(y - 6)}{-(y^2 + y - 42)} \times \dfrac{2(y^2 + 7y + 6)}{y + 1} $ Then factor the quadratic expressions. $x = \dfrac {4(y - 6)} {-(y - 6)(y + 7)} \times \dfrac {2(y + 1)(y + 6)} {y + 1} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {4(y - 6) \times 2(y + 1)(y + 6) } { -(y - 6)(y + 7) \times (y + 1)} $ $x = \dfrac {8(y + 1)(y + 6)(y - 6)} {-(y - 6)(y + 7)(y + 1)} $ Notice that $(y - 6)$ and $(y + 1)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {8(y + 1)(y + 6)\cancel{(y - 6)}} {-\cancel{(y - 6)}(y + 7)(y + 1)} $ We are dividing by $y - 6$ , so $y - 6 \neq 0$ Therefore, $y \neq 6$ $x = \dfrac {8\cancel{(y + 1)}(y + 6)\cancel{(y - 6)}} {-\cancel{(y - 6)}(y + 7)\cancel{(y + 1)}} $ We are dividing by $y + 1$ , so $y + 1 \neq 0$ Therefore, $y \neq -1$ $x = \dfrac {8(y + 6)} {-(y + 7)} $ $ x = \dfrac{-8(y + 6)}{y + 7}; y \neq 6; y \neq -1 $